Log-convexity of the even-index subsequence f(2p) of normalized Eulerian maxima

Determine whether the sequence (f(2p))_{p=2}^∞ is logarithmically convex, i.e., whether f(2p)^2 ≤ f(2p−2) · f(2p+2) holds for all integers p ≥ 2, where f(m) = M_m / m! with M_m = max_{1≤l≤m} A(m,l) and A(m,l) denoting Eulerian numbers; equivalently, f(2p) = J_{2p+1}(1) under the Laplace–Pólya integral correspondence.

Background

Through the identity A(m,l) = m! J_{m+1}(2l − m − 1), the paper relates Eulerian numbers to the Laplace–Pólya integral. Defining f(m) = M_m/m!, with M_m the maximum in the m-th row of Eulerian numbers, yields f(2p) = J_{2p+1}(1) and f(2p+1) = J_{2p+2}(0).

The authors establish monotonicity and convexity properties for subsequences of f(m) and conjecture that the even-indexed subsequence is also log-convex. Current bounds imply a slightly weaker inequality but do not resolve log-convexity.

References

We conjecture that the sequence $\big(f(2p)\big)_{p=2}\infty$ is log-convex as well, although eq:f-ratio-even leads only to the slightly weaker inequality\begin{align*} \frac{f(2p+2)}{f(2p)}\cdot\frac{f(2p-2)}{f(2p)}&\geq \frac{4p3+6p2-2}{4p3+6p2-1}\,. \end{align*}

eq:f-ratio-even:

2p22p2+p−1≤f(2p)f(2p−2)≤2p2+2p−1(p+2)(2p−1).\frac{2p^2}{2p^2+p-1}\leq\frac{f(2p)}{f(2p-2)}\leq\frac{2p^2+2p-1}{(p+2)(2p-1)}.

— Estimates on the decay of the Laplace-Pólya integral (2412.12835 - Ambrus et al., 17 Dec 2024) in Section 5 (Consequences for Eulerian numbers), end of section